COMBINATORICS
If you toss four regular coins, what is the number of possible outcomes? If you roll a
dice how many outcomes are possible? How many different ways can an event go
from one corner of rectangular grid to the diagonally opposite corner? How many
ways can a family of six be arranged in a row to be photographed such that the parents
sit next to each other?
All these questions can be answered using some fundamental principles of counting.
we will discuss various aspects of permutations and combinations. Among others we
will find that Probability Theory is a place where all this knowledge can be applied.
Theorem (The Fundamental Principle of Counting)
Suppose there are k tasks to be performed, of which the first can be done in n1
different ways. For each of those, the second task can be done in n2 different ways and
for each pair of ways in which the first and second task are done there are n3 different
ways the third task can be done etc. and so on. Then the total number of ways in
which the whole of k tasks are performed is n1 n2 n3…nk.
Example
Suppose you roll a die and then toss a coin. The number of ways in which the whole
experiment can be done ( or the number of outcomes for the experiment) is 6  2 = 12.
Example
How many ways can you arrange 5 people in a row? There are five slots, of which the
first can be filled by any of the five people, the second can be filled by any of the
remaining four, etc. So the number of ways is 5  4  3  2  1 = 120.
Definition: n!, pronounced “n factorial”, is defined by n! = 1.2.3........(n-1)n.
Thus the answer in the above example can be written as 5!.
Example
How many 6-letter code words can be made out of all letters and digits if
(a) repetition is allowed?
(b) repetition is not allowed?
Answer: (a) (36)6
(b) 3635  34  33  32  31
Definition: By Pn,r we mean the product of r numbers, starting from n and counting
down. That is,
Pn,r = n(n-1)(n-2)......(n-r+1)
So, for part (b) of the above example, the answer is P36, 6. This P is called the
PERMUTATION symbol.
Example
Out of a group of 10 volleyball players, we need to choose a team of 6. How many
ways can this be done?
Note that the answer to this question is not P10, 6 , since the order in which they are
chosen is unimportant. Suppose the names of the 10 players are A, B, C.I, and J. If the
order were important, ACDHIJ and DCHJAI would have been different, but as it
happens, all of those 6! arrangements of the letters get reduced to just one
combination. This happens to all the combinations, so the answer is P10, 6 /6! =
(10.9.8.7.6.5)/(1.2.3.4.5.6) = 210.
Definition: By Cn,r we mean the number Pn,r /r!
So the answer to the above question is C10, 6.
Note: While Pn,r is the number of ways n objects can be permuted in r places, Cn,r is
the number of ways r objects can be chosen from n objects.
Example
How many ways can you choose 3 objects from a group of 6 objects? The answer is
C6,3 = 20.
Example
If car number plates are such that they consist of 3 digits followed by 3 letters,
repetition permitted, then how many number plates are possible? The answer is
(103)(263).
Example
If in the above problem, 3 letters and three digits can appear anywhere, but repetition
is not permitted, then how many number plates can there be?
The choice of a number plate can be done in 3 steps. First choose 3 places for the
letters to go in, which can be done in C6,3 = 20 different ways. This automatically
selects places for the digits to go in. The second step is to permute a choice of 26
letters in the chosen 3 spots, which can be done in P26,3 = 15600 ways. Finally
permute 10 digits in the remaining 3 spots and we can do this in P10,3 = 720 ways. The
total number of number plates then is (20)(15600)(720) = 224640000.
Note: Different books use different symbols for permutation and combination. Other
commonly used symbols for permutation are (n)r, P and nPr and for combination are
and nCr.
Results:
(i) 0! = 1. (Convention)
(ii) Cn,1 = n and Cn,0 = Cn,n = 1
(iii) Cn,r = Cn,n-r =
n!
r !( n  r )!
(iv) Cn,r = Cn-1,r +Cn-1,r -1
n
(v)  a  b    Cn ,r a r b n r
n
i 0
More Examples
(1) Suppose you toss a coin 10 times. There are 210 possible outcomes. Of these, how
many result in exactly 4 heads? To answer this question all you need to do is to find
how many ways 4 spots could be chosen from 10 spots, so that the heads can go there.
This is clearly C10,4 = 210. In general, for k between 0 and 10, the number of
outcomes with exactly k heads is C10,k.
REVIEW OF SET THEORY
We will leave out the formal definition of a set and let your own idea of what a set is
guide you. Roughly speaking, it is almost any collection of objects, numbers, people,
sets themselves, or anything else that might strike your fancy.
1. The UNION of two sets A and B is the set of all elements that are in A or in B (or
in both), and is denoted by A  B.
2. The INTERSECTION of two sets A and B is the set of all elements that are in
both A and B, and is denoted by A  B
3. A set A is said to be a SUBSET of a set B if all the elements of A are in B. If A is a
subset of B, then B is a SUPERSET of A. We denote this by A  B or B  A. (Some
authors use A  B or B  A.)
4. A UNIVERSAL set is a set that is a superset of all the sets you want to consider.
(When we deal with events in probability theory, we take the sample space to be the
universal set.)
c
5. For a set A, the COMPLEMENT, denoted by A, A
of all elements in the universal set that are not in A.
or A , is the set consisting
6. The EMPTY SET (or the NULL SET), usually denoted by  or { }, is the set with
no elements in it.
7. We say two sets are DISJOINT or MUTUALLY EXCLUSIVE if they have no
common elements, that is, A  B = .
8. The DIFFERENCE between two sets A and B, denoted by A - B or A\B, is
A  B.
USE OF VENN DIAGRAM
Pictorial representation of intersecting and non-intersecting sets are given below
Shaded region is A  B
Three sets, two of them intersecting
Complement of A
A-B
A and B are disjoint
Union of A and B
(AB)c
(AB)c
Such diagrams are called Venn diagrams and are often very useful.
DE MORGAN'S LAWS:
(1)
(2)
A B  A B
A B  A B
RANDOM EXPERIMENTS AND SAMPLE SPACES.
A RANDOM EXPERIMENT or NON-DETERMINISTIC EXPERIMENT is one
whose outcomes are dependent on elements of chance. Rolling a die, tossing a coin
etc. are examples of random experiment. An example of a non-random or
deterministic experiment is dropping a stone from a fixed height and measuring the
time it took to hit the ground. The major difference between the two types of
experiments is that when performed repeatedly, the first type will give different
results, while the second type will give the same result every time, allowing for
experimental errors. It is the randomness that is of primary interest to us.
DEF: The SAMPLE SPACE associated with a random experiment is the set of all
possible outcomes of the experiment. A particular element in the sample space is
called an outcome or a SAMPLE POINT. Subsets of the sample space are called
EVENTS.
EXAMPLE
Roll a pair of dice and observe the pair of numbers that come up. The sample space
for this experiment is given by S = {(1,1), (1,2), (1,3), ...., (5,6), (6,6)}. There are 36
elements in the sample space.
E = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} is an example of an event. In this example,
E is the event that the first die comes up 1. Not every event need to have such simple
interpretation.
Consider the event E1 that the sum of the two numbers that came up is 6. Then E1 =
{(1,5), (2,4), (3,3), (4,2), (5,1)}.
The event that the first die shows up 1 and the sum is 6 is given by EE1 = {(1,5)}.
EXAMPLE
Toss a pair of coins, first of which is double-headed. Then the sample space is given
by S = {HH, HT}. The possible events are , S, {HH}, {HT}. If S has n elements in it,
then the number of events is 2n.
PROBABILITY
DEF: A PROBABILITY is a function that assigns a number P(A) to every event A
such that
(1) P(A)  0
(2) P(S) =1
(3) if A1, A2, A3, ... is an infinite sequence of disjoint events, then

P


1
 
Ai    P  Ai 
 1
RESULTS:
(1) If A1, A2, A3 ,....An is a finite sequence of disjoint events, then
 n  n
P  Ai    P  Ai 
 1  1
(2) P() = 0
(3) If A  B, then P(A)  P(B)
(4) For all A, 0  P(A)  1
(5)
 
P A  1  P  A
DEF: We say that the EQUALLY LIKELY ASSUMPTION holds if the probability
of all outcomes are equal. If the sample space S is a finite set {a1, a2, .....an}, then the
equally likely assumption holds if and only if P{ai} = 1/n for each i.
EXAMPLE:
Roll a pair of dice. What is the probability that the sum is 7? What is the probability
that the sum is 8?
Note that the sample space is {(1,1), (1,2), (1,3), ...., (5,6), (6,6)} with 36 elements in
it and the equally likely assumption holds so that each sample point has probability
1/36. The event that the sum is 7 is {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, so its
probability is 6/36 = 1/6. Similarly, the probability that the sum is 8 is 5/36.
RESULT:
If all the elements in a finite sample space have equal probability, then the probability
of an event E is given by
PE 
Number of elements in E
Number of elements in S
That is,
probability 
Number of favourable outcomes
Total number of outcomes
EXAMPLE:
From a 52-card deck a five-card hand is drawn.
(1) What is the probability that there are no aces in it?
 52 
The sample space has   elements in it and they are all equally likely. Since there
5
 48 
are   possible hands with no aces in them, the probability is
5
 48 
 
5
 52 
 
5
 1712304
2598960 .
(2) What is the probability that there is at least one ace in it?
This event is the complement of the event in (1), so its probability is 1- P(no ace) =
1712304
886656
2598960
2598960
1

(3) What is the probability that there is exactly one ace in it?
 4   48 
Here the number of favourable outcomes is     and hence the probability is
1 4 
 4  48 
  
 1  4 
 52 
 
5
RESULTS:
(1)
P  A  B   P  A  P  B   P  A  B 
(2) If A and B are disjoint, then
(3)
P  A  B   P  A  P  B 

P  A  P  A  B   P A  B
(4) If A  B, then

P  B \ A  P  B   P  A 
EXAMPLE:
A firm is placing three orders for supplies among five distributors. Each order is
randomly assigned to a distributor so a distributor may get multiple orders. Find the
probability that
(a) all orders go to different distributors
(b) all orders go to the same distributor
(c) exactly two orders go to distributor 1.
The answers are: (a)
P5,3
53
(b)
5
53
(c)
 3  4 
  
 2  1 
53
EXAMPLE (THE BIRTHDAY PROBLEM)
Assume that each of 365 days in a year are equally likely to be someone's birthday. If
we have n people in a room, what is the probability that all of them have distinct
birthdays?
Clearly the probability of this event decreases as n increases and finally becomes zero
when n = 366. What is the value of n where the probability goes below 0.5 for the first
time?
P3 6 5,n
The answer to the first question is 365n . To answer the second question, we need to
check the value of the earlier expression for several values of n. It so happens that for
n = 22, the probability is approximately 0.524 and for n= 23 it is approximately 0.493.
DEF: Two events are said to be INDEPENDENT if P(AB) = P(A) P(B).
NOTE: DO NOT confuse independent events with mutually
exclusive events. If two events are mutually exclusive, then P(AB) =
0 instead of P(A) P(B) which is when they are independent.
EXAMPLE:
In the above example A and B are independent, but not disjoint. In the following
example A and B are disjoint but not independent.
NOTE: Two events with positive probability cannot be independent and disjoint at
the same time.
EXAMPLE:
Two coins are tossed. Let A be the event that the first coin comes up head, B be the
event that the second coin comes up tail and C be the event that the second coin
comes up head.
A and B are independent because P(AB) = P{HT} = 0.25 while P(A) P(B) =
(0.5)(0.5) = 0.25. They are clearly not disjoint. Similarly A and C are independent and
are not disjoint.
The events B and C are disjoint and are not independent.
DEF: If A and B are two events with P(B)  0, then the CONDITIONAL
PROBABILITY of A given B, denoted by P(A|B), is defined by
P  A  B
P  A | B 
P  B